Question: The denominator of a fraction is one more than seven times its numerator. If three is added to the numerator and seven is subtracted from the denominator, the resulting fraction is equal to $\frac{5}{8}$. What is the original fraction?
Solution: Let the variables $n$ and $d$ represent the numerator and denominator, respectively. From the problem, we know that \[d=7n+1\] We are also told that \[\frac{n+3}{d-7}=\frac{5}{8}\] We must be careful at this point - it is tempting to think that $n+3$ must equal $5$, but what if the left hand side was not reduced? For example, we might have $n+3=10$ and $d-7=16$ and satisfy the equation. Cross-multiplying the second equation, we see that \[8(n+3)=5(d-7)\] The first equation told us that $d=7n+1$. Substituting this in, we get \[8(n+3)=5(7n+1-7).\] Simplifying, we get \begin{align*}8n+24&=35n-30\\ 27n&=54\\ n&=2\end{align*} Plugging this into our equation for $d$, we get \begin{align*}d&=7n+1\\ d&=7(2)+1\\ d&=15.\end{align*} Thus, our original fraction was $\frac{n}{d}=\boxed{\frac{2}{15}}$.